Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

56
ADVANCED ABSTRACT ALGEBRA
Let x ∈
G k
and g ∈ G
b
Since G G < Z G G
k k+ 1 k+
1
g, we get
Hence Gk , G < Gk+1 and so G k, H < H
bġ
We now get that Gk < N g
H But G k < H
Hence we must have H ≤ N H
G
bg
Corollary : Every maximal subgroup of nil potent G is normal in G.
Proof : Let H be a maximal subgroup of G. Since
hence H∆ G
Theorem 9 :
, by hypothesis we must have
If any finite group G is direct product of its. Sylow subgroups, then G is nilpotent,
Proof : From above theorem 7, it suffices to show that the direct product of two nilpotent groups is nilpotent.
It can be verified easily.
Example 1:
Normal series of Z under addition:
1. l0q∆ 8Z ∆ 4Z ∆Z
and
2. l0q∆
9Z ∆ Z
Examples 2:
l0q∆ 72Z ∆ 8Z ∆Z
can be refined to a series
l0q∆ 72 ∆ 24Z ∆ 8Z ∆4Z∆
Note that two new terms, 24Z and 4Z have been insertd.
Example 3 :
We consider two series of Z 15
:
l0 HN
x
l0q∆
5 ∆ Z15
and l0q∆
3 ∆ Z15
These series are isomorphic
We see that Z
Z
5 Z 3 0
15 5
3 Z 5 0
15 3
≅ ≅ l q
≅ ≅ l q and
Example 4 :
We now find isomorphic refinements of the series given in Example 1
i.e.
(1)
l0q∆
9Z ∆ Z (2)